This is not an elementary solution. I feel like I did something wrong though, as I proved a stronger result for both parts... Note that as $\mathbb{Z}[i]$ is a UFD, it follows $\mathbb{Z}[i][x]$ is a UFD. Now, I will prove a lemma. Suppose $T(x)$ is an irreducible polynomial in $\mathbb{Z}[x]$ such that there exists a polynomial $Q(x)$ such that $T(x)|Q(x)^2 + 1$. Then $T$ splits in $\mathbb{Z}[i][x]$. Suppose not. Remark that $T(x)|(Q(x)+i)(Q(x)-i)$. Then by Euclid's Lemma we have $T(x)|Q(x) \pm i$ for some sign. But this implies $T(x)|i$, which is absurd so $T(x)$ splits (it factors). Now suppose $T(x) = A(x)B(x)$ where $A,B \in \mathbb{Z}[i][x]$ and are nontrivial. Because the norm $N(x) = x\overline{x}$ is multiplicative, we see that $N(A) = N(B) = T(x)$. It follows $A,B$ are irreducible. But conjugating both sides we get $A,B$ are conjugates. Now express $A(x) = M(x) + N(x)i$ where $M,N \in \mathbb{Z}[x]$. Then $T(x) = M(x)^2 + N(x)^2$. Now, express $P(x) = c\pi_1(x)^{e_1}...\pi_n(x)^{e_n}$ where the $\pi_i(x)$ are irreducible in $\mathbb{Z}[x]$ and $c$ is an integer. By above proofs, we have $\pi_i(x) = M_i(x)^2 + N_i(x)^2$ for some $M_i,N_i \in \mathbb{Z}[x]$. Now using the identity $(a^2+b^2)(c^2+d^2) = (ab - cd)^2 + (ad - bc)^2$, we get $\pi_1(x)^{e_1}...\pi_n(x)^{e_n} = A(x)^2 + B(x)^2$ for some $A,B$. a. immediately follows. For b., note that $c=1$ as $P$ is monic so we're done again. EDIT : wait am I being stupid but isn't it true that $c$ can be expressed as a sum of squares as well? So the fact $P$ is monic is useless.

Thanks! Nice solution! Note that in b), we want $A(x)$ and $B(x)$ to have integer coefficients, but if $P(x)$ is not monic, your solution yields to $A(x)$ and $B(x)$ with rational coefficients.

Aha I just caught an error in my proof : I assumed $c$ to be positive. So my proof is actually showing is the leading coefficient is positive then the result holds. This also slightly invalidates my lemma about $T(x) = M(x)^2 + N(x)^2$, but this is easily patched by adding $T$ must have positive leading coefficient. I don't think my solution results in $A,B$ having rational coefficients, because note that $c|Q(x)^2 + 1$ so it follows easily that $c$ or $-c$ is expressible as a sum of squares as it cannot have any prime divisors $3 \pmod{4}$ and then if $P$ has positive leading coefficient it follows it can be written as the sum of two squares.

$\textrm{Any elementary solution?}$